Stuck on these two questions, help me please.

The graph of y=3x^2-4x+2 opens downwards.

True or False?

The graph of y=2x^2-4x+2 has a y-intercept of (0,1).

True or False?

These are properties of a parabola that you absolutely MUST know to even be slightly successful in this topic.

Giving you the answer will not help .

Hint : make a quick sketch to see what the first one looks like

for the second, why not sub in x = 0 as given in the point. Will you get 1 as an answer ???

@reiny

sorry for not replying last night i was tired and didn't realize anyone responded.

for #1 i think it's false?

#2 i think is false
as well the answer will be 2 not 1.

am i right?

To determine whether the graph of y=3x^2-4x+2 opens downwards, we need to analyze the coefficient of x^2.

In the general form of a quadratic equation, y=ax^2+bx+c, the coefficient a determines whether the parabola opens upwards or downwards.

If a is positive, the parabola opens upwards. If a is negative, the parabola opens downwards.

In this case, the coefficient in front of x^2 is 3, which is positive. Therefore, the graph of y=3x^2-4x+2 does not open downwards. The statement is False.

To check if the graph of y=2x^2-4x+2 has a y-intercept of (0,1), we can substitute x=0 into the equation and see if we obtain y=1.

When x=0, we have:

y = 2(0)^2 - 4(0) + 2
y = 0 - 0 + 2
y = 2

The y-coordinate is 2, not 1. Therefore, the y-intercept of the graph is not (0,1). The statement is False.

In summary:

1. The graph of y=3x^2-4x+2 does not open downwards (False).
2. The graph of y=2x^2-4x+2 does not have a y-intercept of (0,1) (False).